If the subspace of all solutions of Ax 0 has a basis consisting of vectors and if A is a matrix, what is the rank of A

Question: If the subspace of all solutions of

Ax = 0

has a basis consisting of vectors and if A is a matrix, what is the rank of A.

Note: The rank of A can only be determined if the dimension of the matrix A is given, and the number of vectors is known. Here in this question, neither the dimension, nor the number of vectors is given.

Assume: The number of vectors is 3, and the dimension is 5 × 8.

The rank of the matrix A is 5.

Step-by-step explanation:

In the standard basis of the linear transformation:

f : R^8 → R^5, x↦Ax

the matrix A is a representation.

and the dimension of kernel of A, written as dim(kerA) is 3.

By the rank-nullity theorem, rank of matrix A is equal to the subtraction of the dimension of the kernel of A from the dimension of R^8.

That is:

rank(A) = dim(R^8) - dim(kerA)

= 8 - 3

= 5

-1   1/6.   you dont subtract the denominators

step-by-step explanation:

6 equilateral triangles